350g ofwater at 30 degree celcius is contained in a copper vessel of mass 50g. Calculate the mass of ice required to bring down the temperature of the vessel and its content at 5 degree celcius.
To calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5 degrees Celsius, we need to determine the amount of heat that will be transferred from the water and the vessel to the ice.
First, let's calculate the heat lost by the water and the vessel. We will use the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q is the amount of heat transferred
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
For the water:
m_water = 350g (given)
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_water = 30°C - 5°C = 25°C
Q_water = m_water * c_water * ΔT_water
For the copper vessel:
m_vessel = 50g (given)
c_copper = 0.39 J/g°C (specific heat capacity of copper)
ΔT_vessel = 30°C - 5°C = 25°C
Q_vessel = m_vessel * c_copper * ΔT_vessel
Now, let's calculate the total amount of heat transferred, which will be equal to the amount of heat gained by the ice:
Q_total = Q_water + Q_vessel
Next, we need to calculate the heat gained by the ice, which can be calculated using the heat of fusion formula:
Q_ice = m_ice * L_fusion
Where:
Q_ice is the amount of heat gained by the ice
m_ice is the mass of the ice
L_fusion is the latent heat of fusion of ice (334 J/g)
Since Q_ice = Q_total, we can equate the two equations:
m_ice * L_fusion = Q_water + Q_vessel
Finally, we can solve for the mass of ice (m_ice):
m_ice = (Q_water + Q_vessel) / L_fusion
Substituting the known values and performing the calculations will give us the required mass of ice to bring down the temperature to 5 degrees Celsius.
To calculate the mass of ice required to bring down the temperature of the water and the copper vessel, we need to consider the heat transfer between them.
The total heat lost by the water and the copper vessel can be calculated using the formula:
Q = mcΔT
Where:
Q is the heat lost or gained
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
First, we need to calculate the heat lost by the water and the copper vessel as they cool down from 30°C to 5°C.
For the water:
m1 = mass of water = 350g
c1 = specific heat capacity of water = 4.18 J/g°C
ΔT1 = change in temperature = 30°C - 5°C = 25°C
For the copper vessel:
m2 = mass of copper vessel = 50g
c2 = specific heat capacity of copper = 0.39 J/g°C (approximate value)
ΔT2 = change in temperature = 30°C - 5°C = 25°C
Now, let's calculate the heat lost by the water and the copper vessel:
Q1 (heat lost by the water) = m1 * c1 * ΔT1
Q1 = 350g * 4.18 J/g°C * 25°C
Q2 (heat lost by the copper vessel) = m2 * c2 * ΔT2
Q2 = 50g * 0.39 J/g°C * 25°C
The total heat lost is the sum of Q1 and Q2:
Total heat lost = Q1 + Q2
Now, let's convert the heat lost to heat gained by the ice as it melts:
Q (total heat lost) = heat gained by the ice
= mass of ice * specific heat capacity of ice * heat of fusion of ice
The specific heat capacity of ice is 2.09 J/g°C and the heat of fusion of ice is 334 J/g.
Therefore:
mass of ice = Q / (specific heat capacity of ice * heat of fusion of ice)
mass of ice = (Q1 + Q2) / (2.09 J/g°C * 334 J/g)
Now, plug in the values and calculate:
mass of ice = (Q1 + Q2) / (2.09 J/g°C * 334 J/g)
mass of ice = (350g * 4.18 J/g°C * 25°C + 50g * 0.39 J/g°C * 25°C) / (2.09 J/g°C * 334 J/g)